Direct Collisions with objects moving in towards each other

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Homework Help Overview

The discussion revolves around a problem involving two smooth spheres of different masses colliding in opposite directions. The smaller mass comes to rest, and participants are tasked with deriving the coefficient of restitution and establishing a relationship between the speeds of the two masses.

Discussion Character

  • Mixed

Approaches and Questions Raised

  • Participants explore the conservation of momentum and the definition of the coefficient of restitution. There are attempts to derive expressions for the coefficient and to understand the implications of the inequality u ≥ 3v.

Discussion Status

Some participants have provided hints and guidance regarding the conservation of momentum and the implications of the coefficient of restitution. There is ongoing exploration of the conditions under which the larger sphere must move after the collision, with various interpretations being discussed.

Contextual Notes

Participants are questioning the assumptions surrounding the directions of motion and the implications of the derived equations. There is a focus on the physical meaning of the coefficient of restitution and its relationship to the speeds of the spheres.

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Homework Statement


Two Smooth spheres of masses m and 3m move with speeds u and v in opposite directions. The smaller mass is brought to rest.

(i) show that the coefficient of restitution,e, is given by e =(u - 3v)/(3u + 3v)

(ii) show that u ≥ 3v

Homework Equations


m1u1 + m2u2 = m1v1 + m2v2

(v1 - v2)/(u1 - u2) = -e


The Attempt at a Solution




(i)
let x equal speed of bigger object after impact

3mv + m(-u) = 3mx

x = (3v -u)/3

using (v1 - v2)/(u1 - u2) = -e


x/(v -(-u) ) = -e

substitute value for x

(3v -u)/(3v + 3u) = -e

e = (u - 3v)/(3v + 3u)

(ii)

can somebody help with this part? I have no Idea how to turn this into an inequality.

Any help would be appreciated
 
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3mv + m(-u) = 3mx
... you have the small sphere traveling in the negative direction before impact and then stopping.
This is a head-on collision: which direction must the larger sphere be traveling after the collision?
What does the speed of the small mass have to have been in order for this to happen?

Now take a look at the equation for the coefficient of restitution ... what happens to it when u < 3v?
 
Simon Bridge said:
... you have the small sphere traveling in the negative direction before impact and then stopping.
This is a head-on collision: which direction must the larger sphere be traveling after the collision?
What does the speed of the small mass have to have been in order for this to happen?

Now take a look at the equation for the coefficient of restitution ... what happens to it when u < 3v?

The larger sphere has to travel in the opposite(negative) direction.

When u < 3v the coefficient of restitution will be negative. (I'm not sure what this means for the impact)

I'm not sure how to work out what speed the small has to be.
 
I'm not sure how to work out what speed the small has to be.
Hint: conservation of momentum.
 
Simon Bridge said:
Hint: conservation of momentum.

Is this correct?

3v - u = 3x

does it have to be u ≥ 3v since we know the bigger mass has to travel in the opposite direction after impact.

which means its magnitude is a negative value. otherwise the bigger mass would continue in the same direction.Which we know can't happen since the smaller one stops.
eg. u = 3.2v 3v - 3.2v = -.2v = 3x

but it could also be equal since they could both stop when they collide.
3v - 3v = 0

hence ......u ≥ 3v
 
does it have to be u ≥ 3v since we know the bigger mass has to travel in the opposite direction after impact.
That's the right idea - I'll just help you tidy up the math a bit:

So - by conservation of momentum: 3v-u=3x => u=3(v-x) (note: u,v>0)
If the small ball stops, the big ball cannot continue forward (the small ball is in the way), so it must stop or go backwards, so x≤0
For x=0, u=?
For x<0, u=?

This can also be obtained from the physical meaning of that e value ... you should take another look at your notes there.
 
Simon Bridge said:
That's the right idea - I'll just help you tidy up the math a bit:

So - by conservation of momentum: 3v-u=3x => u=3(v-x) (note: u,v>0)
If the small ball stops, the big ball cannot continue forward (the small ball is in the way), so it must stop or go backwards, so x≤0
For x=0, u=?
For x<0, u=?

This can also be obtained from the physical meaning of that e value ... you should take another look at your notes there.

Alright thanks
 
Simon Bridge said:
That's the right idea - I'll just help you tidy up the math a bit:

So - by conservation of momentum: 3v-u=3x => u=3(v-x) (note: u,v>0)
If the small ball stops, the big ball cannot continue forward (the small ball is in the way), so it must stop or go backwards, so x≤0
For x=0, u=?
For x<0, u=?

This can also be obtained from the physical meaning of that e value ... you should take another look at your notes there.

x = 0, u = 3v

say x= -1, u =3(v - 1) = 3v - 3 would this not mean ... u < 3v?

or is it because you brought the 3x so it changes from negative to positive
so it would actually be... 3v + 3

and so u > 3v ?
 

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